A `20 Omega` resistor, 1.5 H inductor and `35 mu H` capacitor are connected in series with a 220 V, 50 ac supply. Calculate the impedance of the circuit and also find the current through the circuit.

Explanation

Given R=20 W , L=1.5 H ,
`C=35xx10^(-6)` F
`V_(rms)`=220V ,
f=50 Hz
w.k.t `X_L=2pifL`
=2 x 3.142 x 50 x 1.5
`=471.3 Omega ~~ 91 Omega`
`X_C=1/(2pifC)=1/(2xx3.142xx50xx35xx10^(-6))`
`(X_L-X_C)` = 471.3-91
`=380.3 Omega`
w.k.t.
`Z=sqrt(R^2 + (X_L -X_C)^2)=sqrt(400+144704)`
Impedance , z `~~ 381 Omega`
`therefore I_(rms)=V_(rms)/Z=220/381`=0.578 A
`I_(rms)`=r.m.s value of current = 0.578 A `~~` 0.6 A